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13:00
Can a compact lie group G with dim(G) > 0 have a trival maximal torus?
@Ramanujan 0 is neither (strictly) positive nor (strictly) negative
I don't think so, but I know not enough about Lie theory to decide and can't disprove it.
though the French use a different nomenclature
Actually, whether $0$ belongs to $\mathbb{N}$ or $\mathbb{R}^+$ or not, is a convention... In the majority of countries, it does not. There are exceptions though
thanks then what should i write in domain set for (2^x-1)^1\2 = y @DHMO @Ramanujan our text book says 0 is non negative
13:01
@ffahim that's a rather interesting orientation of the division sign
Oh, and to make a sphere out of a square:: folding it diagonally would work, perhaps? (as in, identify top and left and identify bottom and right)
@ffahim 0 is indeed non-negative
Guys please, nobody familiar with tori in Lie groups in the room / da houz? :(
@DHMO lets say dom : R-{R-} for the function ....AM i write ?
lol
right
you are write
13:04
@ffahim set minus is \
Or you can just say {x|x>=0}
@abenthy ooh pls..... so funny right?
hmmm okay @DHMO thanks broh
@Anubhav First look at the oriented category. $\tilde{M} \to M_1 \# M_2$ be a cover; since $\tilde{M}$ is prime it is irreducible if not $S^1 \times S^2$. Thus $\pi_2(\tilde{M}) = 0$ (otherwise sphere theorem gives an embedded non-null sphere) which would thus imply $\pi_2(M_1 \# M_2) = 0$. I think you should get a counterexample by looking at the separating sphere, but I can't seem to come up with an argument.
user228700
@Ramanujan: U still here?
user228700
(Sorry, I get notifications very late on my phone)
user228700
13:20
Oh, crap did u leave? >.<
user228700
OK, I'll leave the problem here for when u do come back:
@Anubhav Actually, this is easier. Suppose $\tilde{M} \to M$ is a cover and $\tilde{M}$ is irreducible. Then every embedded sphere in $M$ lifts to a sphere in $\tilde{M}$ which is an embedded sphere because $S^2$ has no nontrivial covers. In $\tilde{M}$ the lift bounds a ball, and pushing down gives an embedded ball bounding the original sphere in $M$ (again, the image of that ball is an embedded ball because a ball has no nontrivial cover). This implies $M$ is irreducible.
user228700
I am curious if one can fix the original argument though. It seems to amount to proving that every nullhomotopic sphere in a 3-manifold bounds an embedded ball. I wonder if that's true.
user228700
@Ramanujan: Last question.
user228700
13:26
I was able to figure out that option (A) is correct, but the answer is both (A) and (C).
user228700
Will u please try this and let me know what u get? (U guys finished SL properly, right?)
@Kaumudi can u give me any suggestions that what foreign students' require to get admitted in IIT?
@BalarkaSen this seems work. I was thinking in the line of 1st argument.
Here the basic observation is that covering map sends irreducible into an irreducible
@Anubhav @BalarkaSen can u tell me bro? i am very eager to get admitted in IIT
13:34
I don't know why did I spend so much time on this....I was thinking something absurd and stupid actually
@ffahim No
user228700
@ffahim Uhh, which country are u from?
@ffahim neither of us is an IITan
fro m bangladesh
good, all the best.
13:36
i heard its tough and hard work required but
user228700
@ffahim Which grade are u in?
@Kaumudi Did you remember to take the signs into account when you took the square-roots when computing the distance?
class 10......
user228700
@BalarkaSen I didn't. Squared it; that's probably why I'm missing some roots :/
13:37
@Kaumudi i heard there are some sits for saarc nations..
user228700
@BalarkaSen I won't get anywhere if I don't square it tho :/ Hm, what to do...Wait wait! Ik where I did take square root!
Firstly, the intersection.
ax+by+c=0 --(1)
bx+ay+c=0 --(2)
(1)-(2): (a-b)x-(a-b)y=0
Since a>b, a-b cannot be zero.
Therefore, we can divide a-b from both sides:
x-y=0
x=y
(1): ax+by+c=0
ax+bx+c=0
y=x= -c/(a+b)
---
Then, the distance:
distance = sqrt((1+c/(a+b))^2 + (1+c/(a+b))^2) = sqrt(2(1+c/(a+b))^2)
Since a>b>c>0, (1+c/(a+b)) is positive.
sqrt(x^2) = x when x is positive.
distance = (1+c/(a+b)) sqrt(2) < 2sqrt(2)
1+c/(a+b) < 2
c/(a+b) < 1
c < a+b [a+b is positive]
a+b-c > 0
@Kaumudi I don't see why you'd miss roots if you square though. $x^2 = y^2$ doesn't mean $x = y$.
@Kaumudi However, C is still correct, not because of the distance condition, but because a>b>c>0.
@BalarkaSen in this case it does not matter
Actually, A can be derived also from a>b>c>0, making this question pointless.
It does. (C) works because those are the points on the other side of $(1, 1)$.
user228700
13:41
@DHMO Holy crap, you really needn't have written all this; I already found that (A) is correct! Thanks anyway tho...
"Do you mind stating the statement of the fact that you stated?"... — Paul Plummer 6 mins ago
user228700
@DHMO Yeah, it can. Doesn't make the question pointless tho. Ik that (C) too can be derived from the inequalities given in the question, but I was trying to figure where I went wrong in the distance method.
(A) cannot be derived from a > b > c > 0. It can just be verified from that condition. Typical multiple choice questions sneak.
@BalarkaSen What do you mean?
b>c
b-c>0
a>0
a+b-c>0
user228700
@BalarkaSen From the given condition, we can see that (A) will be correct, no? What's the difference?
13:44
I call this deriving...
That doesn't, at all, prove that distance of those points will be smaller than $2 \sqrt{2}$ when $a + b - c > 0$.
@BalarkaSen No, a+b-c>0 is already a condition.
The question restricted the domain to a>b>c>0, which makes a+b-c>0 true
I interpreted the question as the converse being true (aka if (A)/(C) holds then dist < 2sqrt(2)) too.
if it turns out that no solutions (a,b,c) exist such that the second condition (distance < 2sqrt(2)) is true, we can still say that for all solutions (a,b,c) satisfying both conditions, a+b-c>0 must be true
user228700
@DHMO I don't think that was @BalarkaSen's point. .
13:48
@BalarkaSen This was not stated in the question...
"For a>b>c>0" already restricted the domain
The "then" instead of "It is equivalent to that" also told us that we do not need to account for the converse
@DHMO Fair. Grumble. Typical trick questions.
You win.
@BalarkaSen It's just a discussion. It's not a competition.
Nah, sure. I meant you are correct.
user228700
@BalarkaSen: Hang on. Just. Forget the inequalities. I have $(a+b+c)\sqrt{2}>\sqrt{2}$
What's $d$?
13:50
@Kaumudi Just one rule: You must invert the sign if you are multiplying/dividing by a negative number
user228700
@DHMO Nope. Edited it, sorry.
You should have a factor of $a + b$ on the other side, modulo signs (there should be absolute values too)
user228700
@BalarkaSen Huh?
@Kaumudi How did you derive this?
user228700
Wait. Typo.
user228700
13:52
$(a+b+c)/(a+b)\sqrt{2}>\sqrt{2}$
Da.
Also, absolute values on both sides.
user228700
@BalarkaSen This. Can u explain?
Hello everyone. It has been a while. :)
You need $|\sqrt{x^2}|$, not $x$. Both $x$ and $-x$ are solutions to $z^2 = x^2$. Recall that distance between two points $(a_0, b_0)$ and $(a_1, b_1)$ is $|\sqrt{(a_0 - a_1)^2 + (b_0 - b_1)^2}|$.
@Kaumudi ambiguous division
user228700
13:54
@DHMO M not so great at MathJax. Basically $\sqrt{2}$ should be included with the numerator.
@Kaumudi ok
user228700
@BalarkaSen But the square root of any number is always positive. Wait, wait, so that means that every time I take a square root when there are inequalities involved, I must have that modulus! Correct?
@Kaumudi but $x$ is not always positive
It's called an absolute sign
user228700
@DHMO Yeah, the square root comes with the absolute sign, that's what I meant.
@Kaumudi $\sqrt{x}$ is a solution to $z^2 = x$, and there are two of them. Eg, $z^2 = 4$ has two solutions $2$ and $-2$. You can choose any of them to be $\sqrt{4}$.
user228700
14:00
Yes, yes. OK, but the problem is...
user228700
Posting picture...
I have to chicken out of this conversation now. I am sure someone else can help.
user228700
@BalarkaSen OK, thanks for all the help :-)
user228700
user228700
@DHMO?
14:04
@Kaumudi the first step is wrong
It is like saying $|4|>3\quad\implies\quad-3<4<3$
@DHMO well, technically |4|>3 does imply -3<4<3 :P
@arctictern why?
|u|>c is equivalent to (u<-c or u>c) @Kaumudi, so you did do it wrong
@DHMO because falsehoods imply truths
@arctictern how is |4|>3 false?
err
hides
user228700
14:08
Oh, wait, let me check why I wrote that wrong...
user228700
@Kaumudi that's incorrect
user228700
OK, that explains why I'm getting everything wrong. What's correct?
@Anubhav It follows IIRC from the sphere theorem that if a cover of a 3-manifold is irreducible, then the original 3-manifold is irreducible. In particular, if a cover of a connected sum is prime, then the cover must be prime but not irreducible; it must be $S^2 \times S^1$.
@Kaumudi already said above: |u|>c is equivalent to (u<-c or u>c)
user228700
14:14
@DHMO But that's literally the same thing that I wrote in the picture :/
You wrote -x>a not x<-a @Kaumudi
wait, then that's correct
user228700
@Hippalectryon Right, but then I arrived at that afterward...aren't they basically the same?
uh damn my bad >.> it's correct, just unusual that way
user228700
Phew. I thought I was going mad, being wrong so many times. Alright then, applying this to the original question...
14:17
Is anyone here familiar with hermite polynomials?
I haven't really followed the whole conversation, what's the origina question ? @Kaumudi
In mathematics, the Hermite polynomials are a classical orthogonal polynomial sequence. The polynomials arise in: probability, such as the Edgeworth series; in combinatorics, as an example of an Appell sequence, obeying the umbral calculus; in numerical analysis as Gaussian quadrature; in physics, where they give rise to the eigenstates of the quantum harmonic oscillator; in systems theory in connection with nonlinear operations on Gaussian noise. Hermite polynomials were defined by Laplace (1810) though in scarcely recognizable form, and studied in detail by Chebyshev (1859). Chebyshev's work...
Then it seems like you have to do some amount of annoying work; what free products of 3-manifold groups have $\Bbb Z$ as a finite-index subgroup? It's not obvious to me how to do that, but I'm not very good at group theory.
@0celo7 Just ask and someone might answer
@DHMO in your answer how you get distance = sqrt((1+c/(a+b))^2 + (1+c/(a+b))^2) = sqrt(2(1+c/(a+b))^2)
14:18
@Ramanujan distance formula
user228700
@Hippalectryon Hang on for just one second. Checking...
user228700
Uhhh @DHMO: I still don't get (C):
@DHMO I think you forgot to take LCM
@Ramanujan what?
user228700
14:22
How does one go about proving $$\frac{1}{\sqrt{1-x^2}}\exp\left(-\frac{y^2+z^2-2xyz}{1-x^2}\right)=\exp(-y^2-z‌​^2)\sum_{n=0}^\infty \frac{x^n}{2^nn!}H_n(y)H_n(z)?$$
@MikeMiller I replied to Anubhav's message, I wonder if you saw?
@Kaumudi the second case is automatically rejected because a and b and c are positive
user228700
OK, I don't think we're supposed to arrive at (C) using the given condition for distance; just using the damn inequalities.
Oh,sorry , it looks ([1+c]/a+b)^2 @DHMO
14:23
@Ramanujan ok
user228700
@DHMO Of course.
@Kaumudi already said above
@Kaumudi actually, the second line of the second part is faulty
The first approach turned out to be a dead end, the second is the one.
(though below I wondered if the first approach can be revived)
user228700
14:24
Yeah, I was just trying to figure out if I was doing something wrong, taking the square root.
user228700
@DHMO Huh? Why?
@Kaumudi because you forgot to check if a+b is positive before dividing them from the both sides
user228700
@DHMO I checked; didn't show. We were all aware that I'm not allowed to that if $a+b$ isn't positive, so I didn't show that. It's not like this is an exam, right?
@Kaumudi well, alright
let's move on
user228700
OK, anyhoo, I learned something very valuable from all this. Thanks so much, guys :-)
user228700
14:27
@DHMO Thanks for looking out tho :-)
== Mehler's formula == Mehler (1866) defined a function and showed, in modernized notation, that it can be expanded in terms of Hermite polynomials H(.) based on weight function exp(−x²) as E ( x , y ) = ∑ n = 0 ∞ ( ρ / 2 ) ...
Aha!
@MikeMiller Nice argument!
Thanks.
14:45
All known (simplest) ways to prove 0x=0 for all x:
Of course it relies on Poincare.
@Secret for the first part, why couldn't you use the multiplicative identity on... itself?
What are the axioms?
!!! @Ted studiosus = Moishe Cohen
@DHMO You mean 0=0*1=0?
14:48
@Secret I think I need to know the axioms before I comment
@DHMO I think it is very general, because I am not assuming anything is commutative nor associative
Each arrow adds an axiom
@Secret which axiom did you use for "0=0*1"?
Existence of a right multiplicative identity
@Secret could you state the axiom in symbols?
$\exists e, \forall x \in S : x*e=x$
where S is the algebraic structure
14:51
@Secret where e=1?
in this case, yup, but we do not dismiss the existence of more than one right identity as long there are no left identities
(might need to make this clearer in the slide...)
ok, thanks
@Secret So 0*x=0 is true for all algebraic structure? sorry, new to this
Almost every interesting ones. There are counterexamples such as Wheels
This is because for any known algebraic structure that are interesting, there exists at least one pathway to complete the proof
Wheels are a type of algebra where division is always defined. In particular, division by zero is meaningful. The real numbers can be extended to a wheel, as can any commutative ring. Also the Riemann sphere can be extended to a wheel by adjoining an element 0 / 0 {\displaystyle 0/0} . The Riemann sphere is an extension of the complex plane by an element ∞ {\displaystyle \infty } , where z / 0 = ∞ ...
This?
yes, but even that one, zero has no mutiplicative inverse
Wheel get around that problem by redefining division as a unary operator
the result is you have a lot of zero terms in the expression that would not vanish
14:56
@Secret Could you state id+?
$\exists a, \forall x \in S: a+x=x+a=x$
Thanks
The prefix R or L tells you which of the a+x or x+a you want to discard
If there exists a left and right additive identity, then there is a unique two sided additive identity
@Secret I think you haven't proved (or declared) that equality is reflexive...
Let $a_L$ and $a_R$ be left and right additive identities. Then
$a_L+a_R=a_R$ by properties of $a_L$
$a_L+a_R=a_L$ b properties of $a_R$
Therefore $a_L=a_R$
15:01
@Secret The problem is, you haven't proved that $a_R = a_L + a_R$
I think I already proved that above, since $a_L$ is a left identity, $a_L+x=x$holds for all $x\in S$, and $a_R \in S$ thus it can be one of the $x$
@Secret You have stated that $a_L+x=x$, but not $x=a_L+x$
2 hours ago, by DHMO
Someone said that there is an "aleph fixed point", which means that a number for which the subscript of the aleph is equal to itself. Then, he proceeded to say that that number is an "endless cascade of aleph", meaning $\aleph_{\aleph_{\aleph_{\cdots}}}$. My question is: is this legit?
2 hours ago, by DHMO
2. a torus surface (genus 1) can be formed from a square by identify the two pairs of opposite edges. is there a way to generate a sphere surface (genus 0) with a similar method?
user228700
@ffahim: I'm really sorry, your question got lost. I pretend to be no expert at knowing exactly how to clear the entrance exam, 'cause I haven't been able to do it. And I really have no idea how it's gonna work from next year onward but my only advice would be to Google it. How to do the studying, I mean. It's extremely difficult to manage both school and prep. at the same time and here, where there are coaching institutes that take care of this problem, it's easier for students to do this.
@DHMO I am not sure about the $=$, i think $=$ is defined to be a transitive and reflexive relation, at least that's how most abstract algebra people who wrote papers use it. Otherwise I have no idea how that can be proved
@Secret there are three properties of equality, one of which need to be made axiom
see the 52-line proof of 1+1=2:
here, transitivity is the axiom
(but this is integer arithmetic)
15:07
Ok makes sense. I am guessing it is so widespread that the axiom is often assumed. Anyway, I found a similar MSE post that basically said that same thing as you said:
2
Q: How to prove that equality is an equivalence relation?

PedroProbably, it's a elementary question, but I would like some explanation. Everyone knows that equality relation is (i) reflexive, (ii) symmetric and (iii) transitive, that is, satisfies (i) $x=x$; (ii) $x=y\Rightarrow y=x$; (iii) $x=y$ and $y=z\Rightarrow x=z$. I'm interested in to deduce these...

user228700
@ffahim: And yes, if u plan on preparing, u can rest assured that these will be two of the most difficult years if your whole life, and u'll need people in your life to keep u motivated...
user228700
I can't possibly tell u how to do it right 'cause I've failed at this once already, myself. Do Google it. Also, ask on Quora; there are tons of IITians on that website.
user228700
Best of luck :-)
@Secret alright
I'm pretty sure if I had grown up in India I would be an electrician instead of doing this garbage.
Those exams sound too hard.
15:11
lol
@Secret How would your (simple) diagram look like, if asso. is axiom instead of commutative?
@DHMO That cannot be showed on that diagram, because the arrows there only mean "insert axiom", and not actual binary operations on the boxes
BOD that you meant that as a joke, but be careful.
@Secret I'm talking about a separate diagram
I don't know what BOD means. I did not mean that as a joke.
15:15
@ArtOfCode What's the new avatar?
I do have developed another set of diagrams that do allow associativity and commutativity, it basically look like this:
benefit of doubt
@MikeMiller in that case, be nice
I am being nice - to the users investing a lot of time in something that may not be worth it.
but not to everyone
@MikeMiller Just . . . don't. You mean well. Please mean respectfully well.
15:17
This is somehow the most preposterous discussion I've had on this site.
I don't think it's a reasonable goal for someone to be nice to everyone.
@BalarkaSen nice as in "Nice", as in "respectful, not horrible, rather than actually nice"
Don't be nice, but be respectful.
I've got an echo :)
I/we/whatever don't ask you to coddle people. Just don't be rude.
@ArtOfCode And if I go first, you're the echo. So there.
And who'd complain about such a badass echo?
15:19
That is one of the least rude things I've said in this room. Go find half the discussions with 0celo7 I've had if you want to object to my tone.
In any case, this is worth neither your time nor mine.
Well, true that. Back to Charcoal.
@MikeMiller Being a moderator kinda makes it my job to spend my time on this :)
@MikeMiller What's wrong with your tone in discussions with me?
I really wish someone explained why they dislike me.
Oh well.
@abenthy No. Pick any 1-parameter subgroup; this is isomorphic to $\Bbb R$. Its closure is a compact abelian group, thus a torus. So every positive-dimensional Lie group has a positive-dimensional torus subgroup.
I agree that comment could have been avoided but I, as I am sure as well as many people in this room, are a little annoyed by division-by-zero etc discussions filling up half the chat transcript.
15:22
@Secret it's been 7 minutes...
@Secret in the first arrows of the second section, you forgot to insert id*
@0celo7 Surely you can remember times I've said rather rude things to you. But I've wasted enough time and chatspace talking about metadiscussions.
@DHMO I am still drawing, it takes time to draw things on powerpoint from my notes
@Secret Alright
@MikeMiller No, you behave like most people do towards me...
15:24
I don't understand it
(most people online)
@MikeMiller this is one of the most cringe-worthy videos i have ever seen
the man enjoys drying his hands
you got a problem with that?
@MikeMiller Could I ask you two questions?
15:28
3 hours ago, by DHMO
Someone said that there is an "aleph fixed point", which means that a number for which the subscript of the aleph is equal to itself. Then, he proceeded to say that that number is an "endless cascade of aleph", meaning $\aleph_{\aleph_{\aleph_{\cdots}}}$. My question is: is this legit?
3 hours ago, by DHMO
2. a torus surface (genus 1) can be formed from a square by identify the two pairs of opposite edges. is there a way to generate a sphere surface (genus 0) with a similar method?
Yes, identify two pairs of adjacent edges of a square.
@BalarkaSen nice, thanks
what about genus 2?
You'll get a cone, that's topologically the same as the sphere.
In mathematics, each closed surface in the sense of geometric topology can be constructed from an even-sided oriented polygon, called a fundamental polygon, by pairwise identification of its edges. This construction can be represented as a string of length 2n of n distinct symbols where each symbol appears twice with exponent either +1 or −1. The exponent −1 signifies that the corresponding edge has the orientation opposing the one of the fundamental polygon. == Examples == Sphere: A A − 1 ...
and yes, that thing he describes is a fixed point; more formally there is a function $\aleph$ from cardinals to themselves, and he's describing the union of $\aleph(0)$, $\aleph(\aleph(0))$, etc
@MikeMiller but that looks divergent...
15:33
you need to assume the existence of a weakly inaccessible cardinal, which is a fairly mild assumption that most set theorists (or category theorists, or algebraic geometers, or) are glad to do
I see
in less fancy language you need to prove that there is an "upper bound" on the size of these alephs
once you know that, you can take their union
now iirc this is not something you can prove from the ZF axioms alone but... meh
ok, thanks
How's everyone?
15:38
alive and well
Oooh, someone who remembers MSN.
it's still the thumbs up emote on facebook
Oh. Ohwell.
it looks more like a clapping hand to me
@DHMO here
NB: There is no computational advantage here, it is equivalent to what we are doing with statements
It does, somewhat make computing the associativity cayley table easier by hand because one can easily check what terms to test, and exploit the symmetry of the diagram to find out what terms one can omit to avid double counting
15:44
@Secret How do you prove that $a=b \ \iff\ a+c=b+c$?
I was an AIM guy anyway
@BalarkaSen it takes that text and turns it into a thumbs up; the y means yes
is the inverse axiom necessary?
ah, that's better
guess i shouldn't underestimate facebook
Although people using (y) to mean "Yes" or "Ok" scare me.
(y) feels like "whatever, man."
this is different from the big thumb
15:48
oh, ok.
ah its small and cute.
Is your cover photo 'it's in the syllabus'?
@DHMO Not necessary, as long the structure have the cancellation property (which I do not quite udnerstand as I have not wrap my head around morphisms yet. However for the proofs involving inverses, one can do this as follows:
(NB I forgot to mention one more rule: = can be translated by a pair of arrows with the same value)
@MikeMiller Yes.
@Secret isn't that precisely what i'm trying to prove?
Gauss's original interpretation of Gaussian curvature is/was pretty nice
15:53
You start with a=b, then you add c and conclude a+b=a+c
Meaning?
@Secret isn't that precisely what i'm trying to prove?
Then you start with a+b=a+c, then you add -c (or use cancallation) to show a=b as required
The step by step was not obviosu because I superimpose the two diagrams
@Secret isn't that precisely what i'm trying to prove?
Ok let's get back to written notation: To prove $a=b \ \iff\ a+c=b+c$, you need first show that $a=b \ \Rightarrow\ a+c=b+c$ and then show $a=b \ \Leftarrow\ a+c=b+c$, right?
15:56
@Secret yes
To prove from left to right, you add c on the right, right?
no, that isn't an axiom
Ok I am a bit confused here, what are your axioms, is a,b,c some generic element?

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